Rational Functions

Domain of a rational function

The recipe is short: set the denominator equal to $0$, solve, and exclude those values.

Look at the graph below and notice what happens near $x=-3$ and $x=3$: the curve breaks apart.

Two quick cases to recognize. A repeated factor, like $x^2 - 10x + 25 = (x-5)^2$, still excludes just one value, $x = 5$. And a denominator that is never zero, like $x^2 + 1$, excludes nothing: the domain is all real numbers.

Intercepts, sign, and symmetry

To find the y-intercept, compute $f(0)$ (if $0$ is in the domain). To find the x-intercepts, solve $P(x) = 0$, then check that those values do not also make the denominator $0$. A fraction is zero only when its numerator is zero.

Sign of a rational function

Why? Between two consecutive "special" numbers, neither $P$ nor $Q$ can cross zero, so neither can change sign. One test point tells you the sign for the whole interval.

Even and odd

To test symmetry, replace $x$ with $-x$ and simplify. If $f(-x) = f(x)$, the function is even (symmetric about the y-axis). If $f(-x) = -f(x)$, it is odd (symmetric about the origin). If neither holds, there is no such symmetry.

Vertical asymptotes and holes

When the denominator is $0$ at $x=a$, one of two things happens, depending on whether the numerator is also $0$ there.

When the denominator is a quadratic with an unknown parameter, the discriminant tells you how many zeros it has. If $b^2 - 4ac < 0$, the denominator has no real zeros, so there are no vertical asymptotes (and no holes).

Horizontal and slant asymptotes

Vertical asymptotes tell you what happens near a special $x$. Horizontal and slant asymptotes tell you what happens far away, as $x \to \infty$ and $x \to -\infty$ (the end behavior).

Which side of the slant asymptote?

Writing $f(x) = mx+b+\dfrac{r}{Q(x)}$ also tells you whether the curve lies above or below the asymptote: look at the sign of the leftover fraction $\dfrac{r}{Q(x)}$.

The reciprocal function and transformations

The simplest rational function is the reciprocal function $f(x) = \dfrac{1}{x}$. More generally, $y = \dfrac{k}{x}$ is inverse variation: $k$ is the constant of proportionality, and $xy = k$ always.

Because it is a parent function, it follows the usual transformation rules.

Every rational function of the form $\dfrac{ax+b}{cx+d}$ (with $c\neq0$ and $ad-bc\neq0$) is a transformed reciprocal function. That is why these functions all share the same shape: two branches around a vertical and a horizontal asymptote.

Rational equations, inverses, and compositions

Solving $f(x) = c$

To find where a rational function takes a given value, set up the equation, multiply both sides by the denominator to clear the fraction, solve, and check that the answer is in the domain.

Finding an inverse

To find $f^{-1}$, swap $x$ and $y$ and solve for $y$. With a rational function, the unknown appears twice, so you collect the $y$-terms on one side and factor.

Composing rational functions

To compute $f(f(x))$, substitute the whole expression into itself, then clear the small fractions inside by multiplying the top and bottom by the common denominator.

If you ever find that $f(f(x)) = x$ for every $x$ in the domain, then $f$ undoes itself, so $f = f^{-1}$. Since the graph of an inverse is the reflection of the graph over the line $y=x$, the graph of such a function is symmetric about $y=x$. The simplest example is $y = \dfrac{1}{x}$.

Finding unknown parameters and reading graphs

Sometimes you are given the asymptotes and must find the function. Turn each fact about the graph into an equation.

• Vertical asymptote at $x=a$: the denominator is $0$ at $x = a$ (and the numerator is not).

• Horizontal asymptote $y = L$ with equal degrees: the ratio of leading coefficients is $L$.

• Slant asymptote $y=mx+b$: the quotient of the division is $mx+b$.

• Passes through a point $(x_0, y_0)$: substitute, $f(x_0)=y_0$.

Reading a graph

When you are given a picture, read these in order. First, where does the curve break apart? Those $x$-values are the vertical asymptotes and holes, and they are excluded from the domain. Second, what height does the curve level off to on the far left and right? That is the horizontal asymptote. Third, where does the curve cross the axes? Those are the intercepts. Then build a formula that matches.

For the family $f(x)=\dfrac{ax^2}{x^2+b}$, the numerator and denominator have equal degree, so the horizontal asymptote is $y=a$. The y-intercept is $(0,0)$. If $b>0$ the denominator is never zero: no vertical asymptotes, domain all reals. If $b<0$ there are vertical asymptotes at $x=\pm\sqrt{-b}$. For example, horizontal asymptote $y=2$ with vertical asymptotes $x=\pm3$ gives $a=2$ and $b=-9$, so $f(x)=\dfrac{2x^2}{x^2-9}$.

Rational functions in real situations

Many real quantities are "a fixed amount shared among a changing number": a setup fee shared among hoodies, salt dissolved in a growing volume, a dose spread over time. Each one is a rational function, and its horizontal asymptote has a concrete meaning: the long-term value of the quantity.

Back to the start: the floor for the cost of Ava's hoodies is the horizontal asymptote $y=12$, the price of a single hoodie once the $\$150$ fee is spread thin.

Algebraic tools: bounds, growth, and intersections

Without calculus you can still answer three questions with algebra: "what is the largest value?", "is it always increasing?", and "how many times does a line meet the curve?"

Bounding the outputs with the discriminant

To find which values $y$ a function can take, write $y=f(x)$, clear the fraction, and treat the equation as a quadratic in $x$. The equation has a real solution $x$ only when its discriminant is $\ge 0$. That condition gives the possible values of $y$.

Showing a function is increasing

Divide the polynomials first. A form like $A-\dfrac{B}{\text{something increasing}}$ makes monotonicity easy to see.

Intersecting a line with a rational curve

To find where a line meets a rational curve, set the two expressions equal, clear the denominator, and study the resulting quadratic. The discriminant tells you how many intersection points there are: $\Delta>0$ gives two, $\Delta=0$ gives exactly one, and $\Delta<0$ gives none. Don't forget to throw out any solution that is not in the domain, and to treat the case where the leading coefficient is $0$ separately.

Summary